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DENIUS [597]
3 years ago
11

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

n all directions, how much acoustic power is emitted by the whistle? The threshold of human hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
enyata [817]3 years ago
4 0

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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3 years ago
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A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50
laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

  • Height of the window=1.5 m
  • Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

s=ut+\dfrac{at^2}{2}

Let u be the velocity of the ball when it reaches the top of the window

then

1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

3.96=gt\\t=0.4\ \rm s\\

Let h be the height above the top of the window

h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m

3 0
3 years ago
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